Introduction to The Ideal Gas Law
The Ideal Gas Law summarizes all of the various experiments in the 17th and 18th century that strove to define the relationships between the different properties of a gas. These relationships can all be explained by the ideal gas law. The ideal gas law is based on three critical assumptions.
1. Gas molecules are single points in space and they do not take up any volume, which is a good assumption when the volume of gas molecules is far smaller than the volume of the container.
2. Second, any collisions between gas molecules and other gas molecules, or gas molecules and the surface of the container, are perfectly elastic and do not involve the transfer of energy. The second assumption is sound at lower temperatures near room temperature.
3. The final assumption is that the heat capacity of a gas does not change as a function of temperature, which is a good approximation at lower temperatures but not higher ones.
Gas Laws – Essential Knowledge 2.A.2
Pressure is generated by the collisions of gas molecules against surfaces. These collisions are what cause pressure. Pressure (P) is defined as the average force per unit area on a container. Atmospheric pressure, also known as standard pressure, is the pressure exerted by the air at sea level. It is the defined pressure under which ideal relationships are established. The temperature (T) of a gas is given in units of absolute temperature, the Kelvin scale (after Lord Kelvin). One unit of Kelvin is equal to one degree Celsius \left( 1 \text{ K}={ 1 }^{ \circ } \text{C} \right). The temperature of a gas is directly proportional to the pressure on the container when the volume is held constant, and this relationship is known as the Gay-Lussac law.
This system of holding one variable constant while changing the other two and looking at their relationship was the way all the gas laws were formulated. Robert Boyle published his observation in 1662 that, when holding the temperature of a gas constant, the pressure and the volume of a gas are inversely related to each other (as one goes up, the other goes down). After many experiments and scientific observations over the following hundred years, the collection of rules was compiled in one equation that related all the various properties of a gas. This equation is called the Ideal Gas Law:
PV=nRT, where R=0.08206\dfrac { \text{L atm} }{ \text{mol K} }
Basically put, the product of the volume (V) and pressure (P) of an ideal gas, which are inversely proportional to each other, is equal to the product of the number of gas molecules (n, usually expressed in molar units) times the absolute temperature in Kelvin (T) times a constant of proportionality called the Gas Constant (R). The Gas Constant can be expressed in different units depending on what units the variables are reported on, but its function remains the same: to relate one side of the equation to the other.
This is the keystone equation for gases on the AP® Chemistry exam. All other laws can be derived from this one. Therefore, take the time to understand really what this means. The most common mistake AP® Chemistry students make is trying to solve for a variable without having units that do not match the gas constant. For example, if you wanted to solve for the pressure of a gas, and you were to input the pressure in degrees Celsius instead of Kelvin, your answer would be wrong. This can also happen if you were to input the pressure in units of Pascal instead of atmospheres and forget to change out the gas constant to reflect this change of units. The gas constant can be expressed in many different units, so just make sure that you have selected the version which matches the units in your equation. And don’t forget that sometimes the easiest thing to do could be to convert your units rather than taking the time to find a new value of the gas constant.
We will now provide three examples that showcase questions you may be asked to solve on the AP® Chemistry Exam.
Example 1
What is the volume in liters of a freely expanding container with a pressure of 1 \text{ atm} which contains 1 \text{ mol} of oxygen gas at 25^{ \circ } \text{C}?
The first thing we will do is rearrange the ideal gas law to solve for volume:
V=\dfrac{nRT}{P}
We will then retrieve the value of the gas constant for units of liters and atmospheres. We will use the AP® Chemistry Equation Sheet available on the CollegeBoard website here on page 162.
R=0.08206 \dfrac { \text{L atm} }{ \text{mol K} }
We will then convert the temperature from degrees Celsius to Kelvin.
\text{Temperature (K)}=273+^{\circ}\text{C}=273+25=298 \text{ K}
Finally, we will input all of the variables into the ideal gas law and solve for pressure.
V = \dfrac{(1 \text{ mol})(0.08206\dfrac { \text{L atm} }{ \text{mol K} })( 298 \text{ K})}{( 1 \text{ atm})}
V = 24.45 \text{ L}
This relates to an important point: one mole of an ideal gas occupies 22.4 \text{ L} at 273 \text{ K}. As we know, gases expand when they are heated up. So qualitatively this calculation makes sense — we obtained a result which is a little bit larger than the value at 273 \text{ K}.
Gas Pressure – Essential Knowledge 2.A.2, 4.A.1
According to the ideal gas law, the pressure is related to the number of molecules of a gas. One derivation of this is that the total pressure of a gas is in fact composed of partial pressures of the different molecules that the gas consists of.
{ P }_{ total } = { P }_{ A } + { P }_{ B } + { P }_{ C }...
A gas, like the air we breathe, consists of various constituent molecules of gas such as carbon dioxide (CO_2), nitrogen (N_2), oxygen (O_2), and water (H_2O). Even though the air, in general, has a total pressure, we can measure the contribution of the partial pressure of each type of gas molecule to the total pressure. This relationship is another consequence of the kinetic theory of gases: considering that it is the collisions of each molecule of gas with the wall of the container that provides the total pressure, the amount of pressure exerted by each gas individually is directly proportional to the sum of that particular gas inside the container. This quantity is known as the molar fraction.
{ P }_{ A } = { P }_{ total } \times { X }_{ A }, \text{ where } { X }_{ A } = \dfrac { \text{moles} A }{ \text{total moles} }
Example 2
In a sealed container at room temperature with a fixed volume of 0.5 \text{ L}, 16.5 \text{ grams} of carbon dioxide and 27.0 \text{ grams} of steam are released during a combustion reaction. Assuming that all the oxygen was used up in the reaction, find the partial pressure of carbon dioxide inside the container.
Our ultimate goal is to find the partial pressure of oxygen in the container. We will need to calculate the total number of gaseous moles in the container, plug in that value for n, then use the ideal gas law to solve for the total pressure in the container. Finally, we will use the equation given above for the molar coefficient X_A to find the molar coefficient of oxygen in the container and solve for oxygen’s partial pressure.
The number of moles of a gas is equal to its mass divided by its molecular weight. The molecular weight of carbon dioxide can be calculated by adding the molecular weight of two carbon atoms and one oxygen atoms to equal 44.00 \text{ g}. Steam is gaseous water, and the molecular weight of water is 18.00 \text{ g}. Now let us solve for the number of moles of each compound and add the resulting values together.
\text{Moles} { H }_{ 2 }O\left( { n }_{ water } \right) = 27.0 \text{ g} \times 18.00 \text{ mol/g} =1.50 \text{ mol}
\text{Moles} { CO }_{ 2 }\left( { n }_{ carb } \right) =5.50 \text{ g} \times 44.00\text{ mol/g}= 0.375 \text{ mol}
The total number of moles of gas is the container is therefore.
{ n }_{ total }={ n }_{ water }+{ n }_{ carb }= 1.50\text{ mol} + 0.375\text{ mol} = 1.875\text{ mol}
And the molar fraction of carbon dioxide in the container is:
{ x }_{ carb }=\dfrac { { n }_{ carb } }{ { n }_{ total } } =\dfrac{0.375\text{ mol}}{1.875\text{ mol}}=0.200
Now we arrange the ideal gas equation to solve for the total pressure in the container:
{ P }_{ total }={ n }_{ total }RT{ V }^{ -1 }= (0.375\text{ mol})(0.08206 \text{ L atm} { \text{mol} }^{ -1 }{ \text{K} }^{ -1 })(298 \text{ K})(0.5 \text{ L})
{ P }_{ total }=4.585 \text{ atm}
And finally we solve for the partial pressure of carbon dioxide in the container:
{ P }_{ carb }={ P }_{ total }\times { X }_{ carb }= (4.585\text{ atm})(0.200) = 0.917 \text{ atm}
This type of secondary analysis, of using the ideal gas law to find a value to solve for a second equation, is very common.
Example 3
Natural gas is used to heat up many residential houses. The main component in natural gas is propane. Calculate the number of moles of gas in a 1 \text{ L} container that initially has a pressure of 500 \text{torr}.
The first thing we would need to do in this case is write out and balance the chemical reaction. The combustion of propane will be written out as follows:
{ C }_{ 3 }{ H }_{ 8 }+{ O }_{ 2 }\leftrightarrow C{ O }_{ 2 }+{ H }_{ 2 }O
We can now balance the equation to take into account how many moles of compounds are on each side of the equation. There must be three times as many moles of carbon dioxide as moles of propane since each molecule of propane has three carbon atoms. This reaction will also create four molecules of water for every molecule of propane since propane contains eight hydrogen atoms. The total number of oxygen atoms in three molecules of carbon dioxide and four molecules of water is an even number. Therefore, we do not need to adjust the molar coefficient of propane.
{ C }_{ 3 }{ H }_{ 8 }+5{ O }_{ 2 }\leftrightarrow 3C{ O }_{ 2 }+{ 4H }_{ 2 }O
We would then calculate the number of moles of oxygen in the container using the ideal gas law, rearranged to solve for n. Since solid compounds do not get included in the equilibrium expression, and the question asks for the beginning of the reaction, presumably before the formation of any water or carbon dioxide, the partial pressure of oxygen is just equal to the total pressure of the tank. We will use the ideal gas law flavor of the gas constant in pressure units of torr!
n=\dfrac { RT }{ PV } =(62.36 \text{ L torr} { \text{mol} }^{ -1 }{ \text{K} }^{ -1 })\dfrac { (298 \text{ K}) }{ (500 \text{ torr})(1 \text{ L}) }
n = 37.17\text{ moles} { O }_{ 2 }
There are many other examples of ways the ideal gas law will be tested on the AP® Chemistry exam, and we recommend you try them all! Do you have any questions? Comment below and let us know!
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